6-1 Introduction EXPERIMENT 6 Semiconductors: Preparation of Semiconducting Thin Films Metals are good conductors of electricity. Copper, for example, allows the flow of electrons with relatively little resistance (about 10 6 ohm cm -1 ). It is therefore used in electrical wires. Nonmetals, like glass or diamond, generally have a large resistance to the free flow of electrons (values from 10 10 to 10 18 ohm cm -1 ) and are therefore nonconductors or insulators. Recently, a class of chemical compounds has been developed that are called superconductors, because they conduct electricity with essentially no resistance. Semiconductors have conductivities lying between conductors and insulators (from ~10 8 to ~10 10 ohm cm -1 ). The properties of conductors, semiconductors, insulators, and superconductors depend upon their electronic structures, but it is important to note that the bonding in a bulk structure (like a metal crystal) is quite different from that in a simple molecule. A solid block of a metal consists of a huge collection of atoms. A sample of 6.9 g of lithium metal (one mole of Li) will contain Avogadro's number of atoms (6.0 x 10 23 atoms) and three times Avogadro's number of electrons (each Li atom contains three electrons). Each Li atom is characterized by the electronic configuration ls 2 2s 1 2p 0. In the diatomic lithium molecule, Li 2, the bond between the atoms is caused by the overlap of the valence 2s orbitals, which forms an σ (bonding) and σ* (antibonding) orbital set (see Figure 6-1a, right hand side). Figure 6-1a. Bonding between two atoms with configuration ns 1.
6-2 Each lithium has one valence electron, so the Li 2 molecule has two valence electrons, exactly filling the s bonding molecular orbital (remember each orbital can hold two electrons) and leaving the σ* orbital empty. (In the figure, the 1s 2 electrons are not shown because they do not contribute to bond formation.) The centre of the figure shows how the energy of the molecular orbitals varies with interatomic distance, and on the right are depiction s of the appearance of the electron density and wave function (φ) phase. By considering a series of (hypothetical) longer and longer linear poly-atomic molecules we can understand the band structure of the molecular orbitals in a metals and semiconductors. Figures 6-1b-e show orbital energies and electron distributions for hypothetical linear Li 3, Li 4 and Li 5 molecules. (Notice that the number of molecular orbitals corresponds to the number of atomic orbitals, and that each molecular orbital has one more node than the one below it in energy, from no nodes to n nodes.) Figure 6-1e shows the situation for a very large number of orbitals; so many that they form a virtual continuum of possible energies. This is called a band. In an alkali metal such as lithium half the molecular orbitals would be filled (Figure 6-1f) and metallic conduction would occur because electrons are so easily moved into the empty orbitals. The foregoing considered an unreal situation. Real crystals are three-dimensional and, in addition to the s orbitals, the p orbitals should be considered. Figure 6-1g schematically illustrates these and includes the envelope for the molecular orbitals derived from p overlap. At the atomic separation shown for conductors (usually metals), the band arising from the s orbitals overlaps with the one arising from the p orbitals leading to a continuous conduction band. Figure 6-1b. Bonding in Li 3. Figure 6-1c Bonding in Li 4.
6-3 Figure 6-1d. Bonding in Li 5. Figure 6-1e. Bonding in Li n. Figure 6-1f. Band in L n showing occupied orbitals (darker) Figure 6-1g. Bonding including M.O. s from p orbitals This is not the case for semiconductors (and insulators) which exhibit a band gap. If this gap is small enough, some electrons are thermally promoted from the filled lower band to the empty upper band, and the material is a semiconductor. If the gap is too large for thermal promotion, the element or compound is an insulator.
Consider the bonding found in a real lithium crystal. Suppose that we have n lithium atoms in the crystal and that the crystal is highly symmetric. The individual 2s and 2p orbitals on each lithium can interact with each other in many ways, forming a large number of molecular orbitals of similar energy. These individual molecular orbitals form a nearly continuous series of bands, called the valence band. For lithium, as a result of the high symmetry, the 2s band and 2p band do not exist separately; rather, they overlap and merge into a 2s/2p band. Since there are four energy levels per atom (one 2s and the three 2p orbitals), for n atoms, there will be 4n molecular orbitals, which can hold 8n electrons. Lithium has only one valence electron per atom, however, so there will be only n electrons in the valence band for n atoms. Thus, the valence band is only partly filled, and it is a relatively easy matter to promote an electron from a filled orbital to an empty one, as the energy needed is quite small. In the case of nonmetals like carbon, the structures that are adopted are not so highly symmetric, and the 2s/2p band splits in two, as shown by the line farthest to the right of Figure 6-1g. (This is the equivalent of a crystal forming localized bonding and antibonding orbitals.) The lower energy band is still called the valence band, and the higher energy band is called the conduction band. The energy difference between the bands is called the band gap energy, E g. Since the combined 2s/2p band can hold 8n electrons, each of the split bands can hold half of that, that is, 4n electrons. Since carbon has exactly four bonding electrons per atom, a diamond crystal would have 4n electrons. This number exactly fills the valence band and leaves the conduction band empty. Since the band gap energy is high, free electron flow is prevented. Diamond is therefore classified as an insulator. Silicon and germanium, having the same number of electrons as carbon, might be predicted to be insulators as well. However, they are larger, with less tightly bound electrons and consequently the band gap is relatively small and, particularly as the temperature is raised, conduct electricity weakly thus they are classified as semiconductors. The transition of electrons from the valence band to the conduction band can be achieved by the absorption of electromagnetic radiation, if this energy is equal to or larger than E g. Thus, the absorption of light occurs when: 6-4 hc/λ E g where the wavelength λ is expressed in m (wavelengths are often given in nanometers, nm - do not forget to convert if necessary), h is Planck's constant (6.63 x 10-34 J s), and c is the velocity of light (3.00 x 10 8 m s -l ). If this wavelength happens to fall in the visible region, the semiconductor is known as a photoconductor. Figure 6-2 shows a schematically a portion of the absorption spectrum, and how to obtain the value of E g.
6-5 Figure 6-2. The absorption edge of a photoconductor. The Xerox process depends on this type of conductivity. In a Xerox machine, a positively charged plate is covered by a photoconducting film. Light is reflected by the white part of the original document onto the film, promoting electrons from the valence band to the conduction band, and the film conducts electricity. The parts of the film that were hit by the light are discharged by electrons from the drum on which the film is mounted. The part of the film that was not exposed, corresponding to the black parts of the original, remains charged. Small particles of ink toner are then spread on the film and stick electrostatically only to the charged areas. A new piece of paper (the copy) then removes the toner from the film and thus acquires a copy of the original. Compounds that are isoelectronic (have the same number of electrons) as Si or Ge are also often semiconductors, since they have similar electronic structures. There are two common types-the III/V semiconductors (one element from Group IIIA and one from Group VA, for example, GaAs) and the II/VI semiconductors (one element from Group IIB and one from Group VIA, ZnS or CdS for example). The are other semiconductors which are not isoelectronic and isostructural with Si such as bismuth sulphide. In this experiment, you will prepare a semiconducting film of one of these sulphides (check which one with your demonstrator) on a glass support. Materials: Reagents: 6M ammonium hydroxide solution 0.1M cadmium sulfate solution 0.1 M zinc sulfate solution 0.1M thiourea solution Bismuth(III) nitrate Bi(NO 3 ) 3 Triethanolamine Thioacetamide Ethanol/potassium hydroxide bath
1M Nitric acid Nitrogen gas cylinder for drying slides 6-6 Glassware: 100 ml beaker 3 beakers (10 ml) watchglass Hirsh funnel mortar 1 ml transfer pipette 3 graduated pipette (2 ml) 5 ml graduated pipette Equipment: UV-Vis spectrometer 2 thermometers Stir/Hotplate Tongs Clamps Stand Stirring bar Whatman filter paper (1 cm diameter) Rubber tubing Experimental Procedure NOTE: All glassware, including the glass microscope slides, must be meticulously cleaned. The glassware has to be immersed in a KOH/alcohol bath overnight - this will be done for you. Remove the glassware from the KOH bath using tongs. Rinse the glassware with tap water, followed by immersion in a bath of dilute nitric acid. Rinse again with tap water and finally with deionized water (or distilled water). An ultrasonic cleaning bath containing a detergent solution is an alternative to the KOH bath. After the glass slides are clean and dry, cut them lengthwise into two halves. Your instructor may provide you with the right-sized glass slides. The slides must be cut so that they can easily fit into the spectrophotometer cuvettes. Half of the slide is used for depositing the semiconducting film, and the other half is used as a reference for the absorption spectrum of the film. CAUTION: Take care while handling the cut glass slides. Their edges are sharp
Preparation of a Zinc Sulfide Semiconducting Film 6-7 CAUTION: This reaction generates H 2 S gas, which is toxic. Carry out the deposition procedure inside a well-ventilated hood. Note. The following quantities assume the use of a 10 ml beaker: if a larger one is used, increase the quantities proportionally to obtain a sufficient depth of the mixture to coat the slide to a length of 1-1.5 cm. Set up a hot water bath in a 100 ml beaker on a magnetic stirring hot plate. Place 1.0 ml of 0.1 M zinc sulfate solution in a 10 ml beaker containing a stirring bar. Add 1.0 ml of 0.1 M thiourea (or thioacetamide) solution, followed by 1.5 ml of 6 M NH 3 solution. Heat the beaker in the water bath, and when the temperature has reached 85 o C, begin stirring the solution. At the same time, place the half glass slide halfway into the solution. Maintain the temperature of the water bath at 85 o C. A white precipitate of ZnS will form as the reaction proceeds. Remove the beaker from the water bath after 15 to 20 minutes, using tongs. Slowly and cautiously, remove the slide from the beaker using forceps. Quickly rinse the slide with deionized water. Place the slide in a desiccator for drying under a nitrogen atmosphere. (It can also be dried in a side-arm test tube under vacuum or in a vacuum desiccator - please consult your demonstrator about which procedure to use). These precautions are necessary because zinc sulfide is easily oxidized to zinc sulfate in the presence of moist air. After the slide has dried (20 to 30 min), place it in a cuvette and obtain its absorption spectrum using the other (non-coated) piece as a reference. From the spectrum, determine the value of λ corresponding to the long wavelength limit of the absoption peak (not the maximum), and calculate Eg. Preparation of a Cadmium Sulfide Semiconducting Film CAUTION: Cadmium compounds are toxic. Wear gloves. The H 2 S generated in this procedure is toxic: perform the deposition inside a well-ventilated hood. See the note in the zinc sulphide procedure. Mix 1.0 ml of 0.1 M cadmium sulfate (or cadmium acetate) solution and 1.0 ml of 0.1 M thiourea (or thioacetamide) solution in a 10 ml beaker containing a magnetic stirring bar. Add 2.0 ml of 6 M NH 3 solution. Clamp the beaker in the hot water bath. While stirring the solution, place a half slide halfway into the solution. Maintain the temperature at 70 to 80 o C for 10 to 15 min. During this time a yellow film will deposit on the slide. Remove the slide, rinse it with deionized water, and allow it to air dry. Obtain the absorption spectrum of the semiconductor film, using the other (noncoated) half of the slide as the reference. From the spectrum, determine the value of λ corresponding to the long wavelength limit of the absoption peak (not the maximum), and calculate E g.
6-8 Preparation of a Bismuth Sulfide Semiconducting Film CAUTION: Bismuth compounds are toxic. Wear gloves. The H 2 S generated in this procedure is toxic: perform the deposition inside a well-ventilated hood. See the note in the zinc sulphide procedure. Triturate (pulverize) a mixture of 50 mg of Bi(N0 3 ) 3, 500 µl of triethanolamine (use HOOD), and 2.5 ml of water in a mortar. Using a Hirsch funnel, filter the solution (if necessary) to obtain a clear filtrate. Using a pipet, transfer 1.0 ml of this solution to a 10 ml beaker containing a magnetic stirring bar. Add 1.5 ml of 0.1 M thioacetamide solution, followed by 1.0 ml of 6 M NH 3 solution. Insert a slide halfway into the solution, and place the beaker in a water bath maintained at 85 o C. A brown-black precipitate will soon form, and the slide will be covered with a thin film of Bi 2 S 3. After 10 minutes, remove the slide from the mixture, rinse it with deionized water, and air-dry the slide. Obtain its absorption spectrum as described earlier using the other (noncoated) half of the microscope slide as the reference. From the spectrum, determine the value of λ corresponding to the long wavelength limit of the absoption peak (not the maximum), and calculate E g.
Experiment 6 - Report 6-9 Name Date Results Semiconductor film: Hand in your absorption spectrum Long wavelength limit: (nm) Calculation of E g : Questions 1. What is the purpose of using thiourea or thioacetamide? Write balanced equations for the reactions that occurs when aqueous NH 3 is added to them, and the mixture is heated. 2. ZnS, CdS, and Bi 2 S 3 films absorb in the range of 350, 500, and 950 nm, respectively. In which regions of the electromagnetic spectrum do these absorptions occur? 3. What are n- and p-type conductors? What is meant by doping?